S.Warren. Designing organic syntheses

38

O O O

1,5 - di CO

-

+

O O O O O (best anion)

This sequence can be carried out in one or two steps and makes an important molecule for steroid syntheses. Further details are given in Fleming pages 59, 75 and 171 if you are interested.

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119. If we want to make a simple 1,5-diketone we may have to use an activating group like CO2Et to control the reaction. How would you make TM 119?

OMe

O O

TM 119

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120. Analysis:

	OMe		OMe
			O
O	O	O	+

symmetrical

OMe

O

+

CHO

Synthesis: To ensure good yields, the reaction is best done on an activated compound, so the synthesis becomes:

						O
OMe			O			CO2Et
	O			O		O
				CO 2Et		H+
						TM 119
	base no	MeO		base	MeO	H2O
CHO	ambiguity

Now what about TM 120?

O

TM 120

39

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121. Analysis:

		O		O	O
	O
					CO2Et
		a	a
	O			+ CO2Et	-
			O
					+ i-PrI
		b	add
			control

Choosing the Michael disconnection at a rather than b since we can then use the CO2Et control group both for the alkylation and for the Michael reaction.

Synthesis:

		O			O
O	EtO-		EtO-		1. base
CO2Et				O	TM 120
	i-PrI		O		CO2Et 2. hydrolyse
		CO2Et
					and
					decarboxylate

The final condensation could have gone the other way too, but it doesn't, presumably because attack on, the other carbonyl group is hindered. TM 120 is in fact piperitone, one of the flavouring principles of mint, and has been synthesised essentially by this route (J.C.S.,. 1935, 1583; Rec. Trav. Chim., 1964, 83, 464; Zhur. Obshchei Khim., 1964, 34, 3092, Chem. Abs., 1964, 61, 16098).

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(a)USE OF THE MANNICH REACTION

122. There is one special case worth discussing in some detail. When vinyl ketones (e.g. TM 122) are needed for Michael reactions they may obviously be made by the usual disconnection;

O O

+ CH2O

Which gives formaldehyde as one of the starting materials. Base-catalysed reactions with this very reactive aldehyde often give poor yields because of polymerisation and other side reactions. The Mannich reaction is used instead:

	O				+	O
	+	CH2O +	R2	' NH	H
					R	NR2'
	R

Write a mechanism for this reaction.

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123.

	..			+	..
R2' NH CH2 O H+			R2' N CH2 OH	H	R2' N CH2 OH2+
		..		O
	+	OH
R2	' N CH2	R	R2' N	R	A

40

Alkylation of the product (a 'Mannich Base' A) gives a compound (B) which gives the required vinyl ketone on elimination in base. This last step is usually carried out in the basic medium of tile Michael reaction itself so that the reactive vinyl ketone (TM 122) need never be isolated.

	O	MeI	O	+		base	O
							TM 122
					'	during Michael reacton
R	NR2'		R	NR2			R
	A		B	Me

So how would you make TM 123?

O

Ph

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124. Analysis:

O			O
						O
			Ph	O
Ph	α,β -
		O			O
						+ CH2O
	unsaturated		1,5 - di CO	+
	carbonyl				Ph
				control needed
					(CO2Et)
Synthesis: J. Amer.Chem. Soc., 1954, 76, 4127.
		O
O	base		CO2Et	O	Ph
CO2Et
	PhCH2Br				CO2Et	H+ , H2O
			Ph
				base
						TM 123
O			O
1. CH2O , Me2NH , H+			+NMe3
	2. MeI

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3.REVIEW PROBLEMS

125. Review Problem 9 – Suggest a synthesis of TM 125, a commonly used synthetic intermediate called Hagemann’s ester.

41

CO2Et

TM 125

O

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126. Analysis:

	CO2Et	CO2Et
			CO2Et
	α,β -	1,5 - di CO
	unsaturated		+
	carbonyl	O
		O	O
	O
		O

Synthesis: Though we could follow the stepwise pattern of the disconnections, it is easier to add an activating group to the acetone molecule so that our starting materials are two molecules of acetoacetate and formaldehyde. It turns out that Hagemann's ester can be made in two steps without having to alkylate the Mannich base:

			CO2Et
O	+		+
	CH2O , R2NH , H			1. H / H2O	TM 125
	CO2Et
		EtO2C		2. EtOH / H+

O

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127. Review Problem 10 – Suggest a synthesis for

EtO2C

TM 127	O

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128. Analysls: Disregarding the remote and unhelpful double bond, we can disconnect as a 1,3-dioxygenated compound (frames 94-107).

EtO2C

1,3 - di O

EtO2C

EtO2C

O

Now note symmetry. Doubly allylic disconnection keeps symmetry, requires activation (frames 57-8 and 101-2).

	FGI	EtO2C		EtO2C
Br				+ O
	Br		CO2Et	CO2Et

Synthesis: actually done like this (Chem. Comm., 1967, 753; 1969, 26):

42

			+	-	EtO2C		1. LiAlH4		Br
	O + CH2(CO2Et)2		NH4	AcO
						CO2Et	2. PBr3
	1. NaH , CH2(CO2Et)2		EtO2C			NaH		TM 127
	2. H+ , H2O , heat		EtO2C

3. H+ / EtOH

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129. Review Problem 11 - suggest a synthesis for TM 129

TM 129

O

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130. Analysis: Treat first as an α,β-unsaturated ketone:

				1,5 - di CO
	O	O	O	add
				control
α,β -			α,β -
unsat	+	CH2O	unsat
CO			CO
	O			O

CO2Et

+

O

symmetrical

1,5 - di CO

O

Br

O

A

control

CO2Et

+

O O

Synthesis: All by standard steps. Though the Michael addition on A could in the cry occur at either double bond, the unsubstituted position out of the ring is much more reactive than the disubstituted position in the ring and only the wanted reaction occurs. Bull. Soc. Chim. France, 1955, 8.

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D.ILLOGICAL TWO-GROUP DISCONNECTIONS

1. THE 1,2-DIOXYGENATION PATTERN

(a) α-HYDROXY-CARBONYL COMPOUNDS

131. So far all our two group disconnections have sensible synthons with anions or cations all stabilised by functional groups in the right positions. This won't always be the case. Supposing we wanted to make the hydroxy-acid TM 131; we could treat it as an alcohol:

			H
		O			O	-CO2H
					+
TM 131	Ph		CO2H	Ph
					A

43

but we get the apparently absurd synthon –CO2H. In fact there is a common reagent for this synthon - a simple one-carbon anion which adds to ketones and whose adduct with A could easily be converted into TM 131. What is it?

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132. Cyanide ion! So the synthesis becomes:

O	CN-		OH	NaOH
	+	Ph			TM 131
Ph			CN	H2O
	H

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133. The aldehyde or ketone needed for this reaction is not always readily available. TM 133, labelled with radioactive 14C in one carboxyl group, was needed for a biochemical labelling experiment. How would you make it?

	OH
TM 133	CO2H	= 14C

CO2H

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134. Analysis: The α-hydroxy acid can best be made from an aldehyde and 14CN-, then we can carry on as usual with a 1,3-dicarbonyl disconnection:

OH		CHO
	CN- +		1,3-di CO
CO2H				+
		CO2H	HCO2Et		CO2H
CO2H

Br

+-CH2CO2H

Svnthesis: (J. Amer. Chem. Soc., 1976, 98 , 6380; Tetrahedron, 1972, 28, 1995).

-			-	CHO	-
CH2(CO2Et)2	1. EtO		EtO		1. CN
		CO2Et		CO2Et		TM 133
2.			HCO2Et		2. hydrolysis
	Br

3. etc.

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135. Here is a more difficult example based also on α-hydroxy acids. Use the two phenyl groups as a clue for your first disconnetion in designing a synthesis for TM 135:

	Ph	OH
		OH
	Ph
TM 135
		OH

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136. Analysis: Using the clue, we remove both phenyl groups to give an ester:

44

Ph	OH			-		used in frame 109
	OH	EtO2C	OH	CN +	OH
Ph					OHC
	OH		OH

Synthesis:

	CH2O		1. CN-			1. EtOH / H+
		OH	HO2C		OH		TM 135
CHO	K2CO3	OHC	2. HO-/ H2O	OH		2. PhMgBr
					A

We can either protect the two hydroxyl groups in A as a cyclic acetal or use four mols of PhMgBr and waste two of them.

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137. A more elaborate variation gives a generell amino acid synthesis. If the reaction between an aldehyde and cyanide is done in the presence of ammonia, the product is an α-amino-nitrile:

NH3 , CN- NH2

RCHO

RCH CN

Can you see what intermediate is being trapped by the cyanide ion?

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138. It must be the imine:

	NH3	NH2
RCHO	RCH NH	RCH CN
	-CN	A

NH2

RCH CO2H

B

Under the right conditions, hydrolysis of the cyanide A occurs during the reaction to give the amino acid B. How could you make the amino acid Valine (TM 138)?

	TM 138	HO2C	NH2
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139. Analysis:
	NC-		+ NH3
HO2C	NH2	NH	CHO
Synthesis:		NH3 , CN-
	CHO		TM 138

This is the Strecker amino acid synthesis.

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45

140. Strangely enough, cyanide ion is also involved in one special reaction giving an α-hydroxy-ketone. Can you show how the adduct A of' benzaldehyde and cyanide ion can give a stable 'carbanion'?

				Ph	Ph
			A	O	OH
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141.
		Ph	OH			Ph	OH
		B : H	C				C
				N			N -
This anion	now	reacts with another molecule of				benzaldehyde		to give	eventually the
α-hydroxy-ketone 141A. Draw mechanisms for these steps:
Ph	OH						Ph	Ph	CN-
	C	+ PhCHO					O	+
	N -							OH
							141A
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142.
Ph	OH	Ph			Ph	Ph
	C	C				OH		141A
				H	O
- N		O
					CN

The product is called benzoin and the reaction is known therefore as the benzoin condensation. No base is needed other than cyanide ion.

CN	-	Ph	Ph
PhCHO			benzoin
		O	OH

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143. How could benzoin be elaborated into the more complex molecule TM 143?

		Ph
		Ph
		O
	TM 143	Ph
		Ph

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144. Analysis: We can disconnect both the symmetrical α,β-unsaturated carbonyl linkages:

46

Ph		Ph			Ph
Ph	Ph	O		FGI	OH	HCO2Et + PhCH2MgBr
O		O	O
Ph	Ph				Ph
Ph		Ph
Synthesis:
CN-	Ph	Ph	CrO3	Ph	Ph
PhCHO
	O	OH		O	O
						base
		Ph			Ph	TM 143
1. Mg , Et2O			CrO3
PhCH2Br		OH			O
2. HCO2Et
		Ph			Ph

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145. The same problem of illogicality arises with other α-hydroxyketones:

H

O	-
O	+
	O
O	A

Again we need a reagent for an acyl anion synthon (A). We find this in the acetylide ion since substituted acetylenes can be hydrated to ketones:

Hg2+ , H+ O

R H R

H2O

If you want to know more about this reaction, see Norman p.116 or Tedder, vol 1, p.108.

TM 145

How then could one make TM 145?

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146. Synthesis:

1. Na , lig NH3

H H

2. acetone

OH

O

OH Hg2+ , H+

TM 145

H2O

The reaction can be used for disubstituted acetylenes, but it is unambiguous only when they are symmetrical. Suggest a synthesis for TM 146.

O

TM 146

O

47

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147. Analysis: The	cyclic ether	is obviously	made from a diol, and	that gives us a
1,2-dioxygenated skeleton of the right kind:
O	O
FGI		FGI	O + H	H + O
	OH HO
O		HO	OH

Synthesis: We need the symmetrical double adduct from acetone and acetylene.

				1. base ,	O
H				H
				2. base ,	O

Hg2+ , H+

TM 146

HO

OH

H2O

The ether forms spontaneously from the tertiary alcohols in acid.

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148. α-Hydroxy ketones take part in condensation reactions too. How would you make TM 148?

O

OH

TM 148

O

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149. Analysis: Start with the α,β-unsaturated relationship as the alternative (the 1,2-di O) is no good at the start. After the first disconnection we have a methyl ketone which can come from an acetylene:

		CHO +	FGI
			OH
O	OH	O	OH
	O		O

H H +

O

Synthesis: (Ber. , 1922, 55, 2903 for the later stages).

1. Na , lig NH3

H H

2. acetone

OH

Hg2+ , H+

H2O

O OH

CHO

O

readily available furfural

TM 148

91%